It's a typing mistake (in the last line). It should be $7^{78}$.

2 votes

**Option D) is correct**

The Cyclicity table for $7$ is as follows:

$7^1 =7$

$7^2 =49$

$7^3 = 343$

$7^4 = 2401$

Hence cyclicity of $7$ is $4$

Let’s divide $78$ by $4$ and the remainder is $2$

Thus, the last digit of $7^78$ is equal to the last digit of $7^2$ i.e. $9$

2 votes

To get the unit digit of $7^{78}$, we will find $7^{78}\;mod\;10.$

$7^{78}\; mod\;10=7^{76+2}\;mod\;10$

$=(7^{76}*7^{2})\;mod\;10$

$=(7^{76}\;mod\;10*7^{2}\;mod\;10)\;mod\;10$

$=((7^{4})^{19}\;mod\;10*7^{2}\;mod\;10)\;mod\;10$

$=(1*7^{2}\;mod\;10)\;mod\;10$ ( $\because$ when $a,n$ are co-primes then $a^{\Phi (n)}mod\;n=1$. Here $7,10$ are co-primes and $\Phi (10)=4$. So, $7^{4}\;mod\;10=1$ ) $[\Phi(n) = Euler \;\;Totient\;\; Function ]$

$=(49\;mod\;10)\;mod 10$

$=9\;mod\;10$

$= 9$

$7^{78}\; mod\;10=7^{76+2}\;mod\;10$

$=(7^{76}*7^{2})\;mod\;10$

$=(7^{76}\;mod\;10*7^{2}\;mod\;10)\;mod\;10$

$=((7^{4})^{19}\;mod\;10*7^{2}\;mod\;10)\;mod\;10$

$=(1*7^{2}\;mod\;10)\;mod\;10$ ( $\because$ when $a,n$ are co-primes then $a^{\Phi (n)}mod\;n=1$. Here $7,10$ are co-primes and $\Phi (10)=4$. So, $7^{4}\;mod\;10=1$ ) $[\Phi(n) = Euler \;\;Totient\;\; Function ]$

$=(49\;mod\;10)\;mod 10$

$=9\;mod\;10$

$= 9$

1

You wrote this line. But this is NOT always true. For example, $8$ and $15$ are co-prime, but $8^{14} \mathrm{~mod~} 15 = 4 \ne 1$.

1

Thanks for pointing it out.. I 've hidden the answer for now.. I' ll check and update it later..

Ps:@techbd123 can you check now...

Ps:@techbd123 can you check now...